GCSE WJEC Chemistry: pH Calculations – Key Points | GCSE WJEC 化学：pH计算考点精讲

📚 GCSE WJEC Chemistry: pH Calculations – Key Points | GCSE WJEC 化学：pH计算考点精讲

The pH scale is a fundamental concept in chemistry that measures how acidic or alkaline a solution is. Mastering pH calculations is essential for success in the WJEC GCSE Chemistry exam, as questions often involve strong acids, strong bases, dilution, and neutralisation. This guide breaks down every core topic you need to know, from the definition of pH to mixed-solution calculations, using clear examples and the required formulae.

pH 标度是化学中衡量溶液酸性或碱性强弱的基本概念。掌握 pH 计算对于 WJEC GCSE 化学考试至关重要，因为试题常涉及强酸、强碱、稀释和中和反应。本指南将你需要掌握的每个核心知识点逐一拆解，从 pH 的定义到混合溶液计算，配合清晰的示例和必需的公式，助你稳拿高分。

1. The pH Scale | pH 值标度

The pH scale runs from 0 to 14 and indicates the concentration of hydrogen ions, H⁺, in a solution. A pH less than 7 is acidic, pH = 7 is neutral, and pH greater than 7 is alkaline. The scale is logarithmic, meaning each whole pH change corresponds to a tenfold change in H⁺ concentration.

pH 标度范围是 0 到 14，表示溶液中氢离子 H⁺ 的浓度。pH 小于 7 为酸性，等于 7 为中性，大于 7 为碱性。该标度是对数关系，即 pH 值每变化 1 个单位，H⁺ 浓度就变化 10 倍。

2. The Key Equations for pH | pH 计算核心公式

WJEC requires you to memorise and apply two fundamental equations for strong acids and bases at 25 °C (298 K):

WJEC 考试要求你熟记并运用两个适用于 25 °C (298 K) 下强酸和强碱的基本公式：

pH = -log₁₀[H⁺]

[H⁺] = 10⁻ᵖᴴ

Kw = [H⁺][OH⁻] = 1 × 10⁻¹⁴ mol² dm⁻⁶

The square brackets denote concentration in mol dm⁻³. The ionic product of water, Kw, links H⁺ and OH⁻ concentrations and is essential for base calculations.

方括号表示浓度，单位是 mol dm⁻³。水的离子积 Kw 将 H⁺ 与 OH⁻ 的浓度联系起来，是计算碱溶液 pH 的关键。

3. Calculating the pH of Strong Monoprotic Acids | 强一元酸的 pH 计算

Strong acids like HCl, HNO₃ and H₂SO₄ dissociate completely in water. For a monoprotic acid such as HCl, the concentration of H⁺ equals the acid concentration. If you have 0.0100 mol dm⁻³ HCl, then [H⁺] = 0.0100 mol dm⁻³. pH = -log₁₀(0.0100) = 2.00.

强酸（如 HCl、HNO₃ 和 H₂SO₄）在水中完全电离。对于 HCl 这样的一元酸，H⁺ 浓度等于酸的浓度。如果 HCl 浓度为 0.0100 mol dm⁻³，则 [H⁺] = 0.0100 mol dm⁻³，pH = -log₁₀(0.0100) = 2.00。

4. Calculating the pH of Strong Diprotic Acids | 强二元酸的 pH 计算

Sulfuric acid, H₂SO₄, releases two H⁺ ions per molecule upon complete dissociation. Therefore, [H⁺] = 2 × acid concentration. For 0.00500 mol dm⁻³ H₂SO₄, [H⁺] = 0.0100 mol dm⁻³, giving pH = 2.00. Always check whether the acid is mono- or diprotic.

硫酸 H₂SO₄ 完全电离时每个分子释放出两个 H⁺。因此 [H⁺] = 2 × 酸的浓度。0.00500 mol dm⁻³ 的 H₂SO₄ 对应的 [H⁺] = 0.0100 mol dm⁻³，pH 同样为 2.00。务必先判断酸是一元还是二元酸。

5. Calculating the pH of Strong Bases Using Kw | 利用 Kw 计算强碱的 pH

Strong bases like NaOH and KOH fully dissociate to give OH⁻ ions. To find pH, first calculate [OH⁻] from the base concentration, then use Kw to find [H⁺]:

强碱（如 NaOH 和 KOH）完全电离产生 OH⁻ 离子。要计算 pH，首先根据碱的浓度求出 [OH⁻]，再通过 Kw 计算 [H⁺]：

[H⁺] = Kw / [OH⁻] = 1 × 10⁻¹⁴ / [OH⁻]

Then apply pH = -log₁₀[H⁺]. For example, 0.100 mol dm⁻³ NaOH gives [OH⁻] = 0.100 mol dm⁻³, [H⁺] = 1 × 10⁻¹³ mol dm⁻³, so pH = 13.00. For a diprotic base like Ba(OH)₂, [OH⁻] = 2 × base concentration.

然后用 pH = -log₁₀[H⁺] 计算。例如 0.100 mol dm⁻³ NaOH 得到 [OH⁻] = 0.100 mol dm⁻³，[H⁺] = 1 × 10⁻¹³ mol dm⁻³，pH = 13.00。对于二元碱如 Ba(OH)₂，[OH⁻] = 2 × 碱的浓度。

6. Effect of Dilution on pH of Strong Acids | 稀释对强酸 pH 的影响

Diluting a strong acid by a factor of 10 increases the pH by 1, because [H⁺] falls to one‑tenth of its original value. For instance, diluting a pH 2 HCl solution (0.01 mol dm⁻³) ten times yields a 0.001 mol dm⁻³ solution, giving pH = 3. This simple rule only applies to strong acids where dissociation remains complete.

将强酸稀释 10 倍，pH 值增加 1，因为 [H⁺] 降低为原来的十分之一。例如将 pH = 2 的 HCl 溶液（0.01 mol dm⁻³）稀释 10 倍，浓度变为 0.001 mol dm⁻³，pH = 3。这个简单规律仅适用于完全电离的强酸。

7. Effect of Dilution on pH of Strong Bases | 稀释对强碱 pH 的影响

When a strong base is diluted 10 times, [OH⁻] decreases by a factor of 10, so [H⁺] increases tenfold and the pH falls by 1. For example, a pH 11 NaOH solution (0.001 mol dm⁻³) diluted tenfold gives pH 10. Remember, dilution always pulls the pH towards 7 but never crosses it for strong acids or bases at ordinary concentrations.

强碱溶液被稀释 10 倍时，[OH⁻] 减小为原来的十分之一，因此 [H⁺] 增加 10 倍，pH 降低 1。例如 pH = 11 的 NaOH 溶液（0.001 mol dm⁻³）稀释 10 倍后 pH = 10。请记住，稀释总是使 pH 向 7 靠近，但在常规浓度下强酸或强碱的 pH 不会越过 7。

8. pH of Mixtures After Neutralisation | 中和反应后混合溶液的 pH

When an acid and a base are mixed, neutralisation occurs: H⁺ + OH⁻ → H₂O. To calculate the pH of the resulting solution, first determine which reactant is in excess. Find the moles of H⁺ and OH⁻ before mixing, subtract the smaller number from the larger, then find the final concentration of the excess ion in the total volume. Finally, compute [H⁺] and pH.

当酸与碱混合时发生中和：H⁺ + OH⁻ → H₂O。要计算所得溶液的 pH，首先判断哪种反应物过量。求出混合前 H⁺ 和 OH⁻ 的物质的量，用大值减去小值，再根据总体积计算过量离子的最终浓度，最后求出 [H⁺] 和 pH。

Example: 25.0 cm³ of 0.100 mol dm⁻³ HCl is added to 30.0 cm³ of 0.100 mol dm⁻³ NaOH. Moles HCl = 0.00250, moles NaOH = 0.00300. Excess OH⁻ = 0.000500 mol in 55.0 cm³ (0.0550 dm³), so [OH⁻] = 0.00909 mol dm⁻³. Using Kw, [H⁺] = 1.10 × 10⁻¹² mol dm⁻³, pH = 11.96.

示例：将 25.0 cm³ 0.100 mol dm⁻³ HCl 加入 30.0 cm³ 0.100 mol dm⁻³ NaOH 中。HCl 的物质的量是 0.00250 mol，NaOH 是 0.00300 mol。过量 OH⁻ 为 0.000500 mol，总体积 55.0 cm³ (0.0550 dm³)，[OH⁻] = 0.00909 mol dm⁻³。利用 Kw 得 [H⁺] = 1.10 × 10⁻¹² mol dm⁻³，pH = 11.96。

9. Calculating [H⁺] from pH | 已知 pH 求 [H⁺]

To convert a pH value back to hydrogen ion concentration, use [H⁺] = 10⁻ᵖᴴ. This is often required when comparing acid strengths or when given a pH and needing the original concentration. For example, a solution with pH 4.30 has [H⁺] = 10⁻⁴·³⁰ = 5.0 × 10⁻⁵ mol dm⁻³.

若要将 pH 值转换回氢离子浓度，使用 [H⁺] = 10⁻ᵖᴴ。这在比较酸强度或已知 pH 求原始浓度时经常用到。例如 pH 为 4.30 的溶液，[H⁺] = 10⁻⁴·³⁰ = 5.0 × 10⁻⁵ mol dm⁻³。

10. Common Pitfalls and Exam Tips | 常见易错点与考试技巧

Always write the full unrounded value from your calculator before rounding to the appropriate number of decimal places – pH values are typically given to 2 decimal places. Remember that [H⁺] = 10⁻ᵖᴴ gives you the concentration in mol dm⁻³. For diprotic acids, remember to multiply the acid concentration by 2 to get [H⁺]. Never forget to use Kw when working with bases. Watch units: convert cm³ to dm³ (÷1000) before calculating moles.

在计算器上保留未舍入的完整数值，再按照要求（通常 pH 保留两位小数）进行舍入。记住 [H⁺] = 10⁻ᵖᴴ 得到的是以 mol dm⁻³ 为单位的浓度。对于二元酸，记得将酸的浓度乘以 2 得到 [H⁺]。处理碱时一定不要忘记使用 Kw。注意单位：计算物质的量前将 cm³ 转换为 dm³（除以 1000）。

pH is a log scale: a change of 1 means ×10 change in [H⁺].
稀释时 pH 每变化 1，[H⁺] 变化 10 倍。
Always identify the excess reactant in neutralisation mixtures.
中和反应混合物中务必先找出过量反应物。
State symbols and significant figures matter in structured questions.
结构化试题中状态符号和有效数字同样重要。

11. Quick Reference Table of Common Acids and Bases | 常见酸碱速查表

Solution	Concentration / mol dm⁻³	[H⁺] or [OH⁻]	pH
HCl	0.1	[H⁺] = 0.1	1.00
H₂SO₄	0.05	[H⁺] = 0.1	1.00
NaOH	0.1	[OH⁻] = 0.1	13.00
Ba(OH)₂	0.05	[OH⁻] = 0.1	13.00
Pure water	—	[H⁺] = 1×10⁻⁷	7.00

12. Practice Question Walkthrough | 典型例题精析

Question: 20.0 cm³ of 0.200 mol dm⁻³ HCl is mixed with 15.0 cm³ of 0.100 mol dm⁻³ NaOH. Calculate the pH of the resulting solution.

题目：将 20.0 cm³ 0.200 mol dm⁻³ 的 HCl 与 15.0 cm³ 0.100 mol dm⁻³ 的 NaOH 混合。计算所得溶液的 pH。

Solution / 解答：

Moles HCl = (20.0/1000) × 0.200 = 0.00400 mol, so [H⁺] from acid = 0.00400 mol. Moles NaOH = (15.0/1000) × 0.100 = 0.00150 mol, so [OH⁻] = 0.00150 mol. Excess H⁺ = 0.00400 – 0.00150 = 0.00250 mol. Total volume = 35.0 cm³ = 0.0350 dm³. [H⁺] = 0.00250 / 0.0350 = 0.0714 mol dm⁻³. pH = -log₁₀(0.0714) = 1.15.

HCl 的物质的量 = (20.0/1000) × 0.200 = 0.00400 mol，即酸提供的 H⁺ 为 0.00400 mol。NaOH 的物质的量 = (15.0/1000) × 0.100 = 0.00150 mol，OH⁻ 为 0.00150 mol。过量 H⁺ = 0.00400 – 0.00150 = 0.00250 mol。总体积为 35.0 cm³ = 0.0350 dm³。[H⁺] = 0.00250 / 0.0350 = 0.0714 mol dm⁻³。pH = -log₁₀(0.0714) = 1.15。

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GCSE WJEC Chemistry: pH Calculations – Key Points | GCSE WJEC 化学：pH计算 考点精讲